If each term of a geometric progression a1, a2, a3, ... with a1 = 1/8 and a2 ≠ a1, is the arithmetic

Q: Why can't we just use the standard sum formula $S_n = \frac{a(r^n-1)}{r-1}$?

Using the sum formula is mathematically valid, but it is computationally exhausting and prone to error. In JEE Advanced, we look for elegance. Subtracting the sums directly reveals the beautiful cancellation $S_{20} - S_{18} = a_{19} + a_{20}$. Always look for the shortcut before diving into the heavy algebra.

Q: Why do we reject $r=1$?

The problem explicitly states $a_2 \neq a_1$. If $r=1$, then $a_2 = a_1 \cdot 1 = a_1$, which contradicts the given condition. Always check your constraints at the start; they are often the key to filtering out extraneous solutions.

Q: How do I handle the negative sign in $(-2)^{18}$?

Remember that any negative number raised to an even power becomes positive. So, $(-2)^{18} = 2^{18}$. This is a common place to lose marks, so stay sharp and keep your signs organized.

Question

If each term of a geometric progression $a_1, a_2, a_3, \dots$ with $a_1 = \frac{1}{8}$ and $a_2 
eq a_1$, is the arithmetic mean of the next two terms and $S_n = a_1 + a_2 + \dots + a_n$, then $S_{20} - S_{18}$ is equal to

Options:

(A) $2^{15}$

(B) $-2^{18}$

(C) $2^{18}$

(D) $-2^{15}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a math problem; we are uncovering a hidden symmetry within a geometric progression.

When you look at a sequence like $a_1, a_2, a_3, \dots$, it is easy to get lost in the notation. But I want you to see the rhythm. We are given that each term is the arithmetic mean of the next two, which is a powerful constraint governing the sequence's internal balance.

### Decoding the Condition

Let us translate the prose into the language of the universe: mathematics. The problem states that:

$$a_n = \frac{a_{n+1} + a_{n+2}}{2}$$

Multiplying both sides by $2$, we obtain the balance equation:

$$2a_n = a_{n+1} + a_{n+2}$$

Now, we apply the definition of a geometric progression, where any term $a_n$ is written as $ar^{n-1}$. Substituting this into our balance equation, we get:

$$2(ar^{n-1}) = ar^n + ar^{n+1}$$

Since $a_1 = \frac{1}{8}$, we know $a 
eq 0$. We can safely divide by $ar^{n-1}$ to simplify the expression into an elegant quadratic:

$$2 = r + r^2$$

### The Quadratic Trap

Rearranging the equation, we get $r^2 + r - 2 = 0$. Factoring this is straightforward:

$$(r + 2)(r - 1) = 0$$

This gives us two potential paths: $r = -2$ or $r = 1$. However, the problem explicitly states $a_2 
eq a_1$.

If $r$ were $1$, every term would be identical, making $a_2 = a_1$. Therefore, we must reject $r = 1$. Our common ratio is locked in as **$r = -2$**.

### The Final Sprint

We need to find $S_{20} - S_{18}$. Many students would immediately reach for the sum formula, but that is a trap.

Visualize the sums: $S_{20} = a_1 + a_2 + \dots + a_{18} + a_{19} + a_{20}$ and $S_{18} = a_1 + a_2 + \dots + a_{18}$. When you subtract $S_{18}$ from $S_{20}$, everything from the first term to the eighteenth term cancels out perfectly.

We are left with only $a_{19} + a_{20}$. Using our G.P. formula, this is $ar^{18} + ar^{19}$, which factors to:

$$ar^{18}(1 + r)$$

Substituting our known values, $a = \frac{1}{8}$ and $r = -2$, we get:

$$\left(\frac{1}{8}ight)(-2)^{18}(1 - 2)$$

Since $8 = 2^3$, this becomes $(2^{-3})(2^{18})(-1)$. Using the laws of exponents, we add the powers:

$$-2^{18-3} = -2^{15}$$

The final result is **$-2^{15}$**. You have mastered the sequence.

Select Answer:

Visualized Solution

Understanding the G.P.

The Arithmetic Mean Condition

Substituting G.P. Terms

Simplifying the Equation

Forming the Quadratic Equation

Solving for Common Ratio $r$

Filtering the Value of $r$

Defining the Target Expression

Expressing in terms of $a$ and $r$

Substituting Known Values

Final Calculation

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

Decoding the Condition

The Quadratic Trap

The Final Sprint

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